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How to replace string values with "XXXXX" in java?

By : fyzm7357
Date : July 29 2020, 02:00 PM
it fixes the issue I want to replace particular string values with "XXXX". The issue is the pattern is very dynamic and it won't have a fixed pattern in input data. , This may cover given both cases.
code :
String input = "https://internetbanking.abc.co.uk/personal/logon/login/?userId=Templ3108&password=Got1&reme";

// Mask UserId
System.out.println(input.replaceAll("userId=(\\w+)", "userId=XXXXX"));

// Mask Password
System.out.println(input.replaceAll("password=(\\w+)", "password=XXXXX"));
String maskUserNameAndPassword(String input) {
    return input.replaceAll("userId=(\\w+)", "userId=XXXXX")
                .replaceAll("password=(\\w+)", "password=XXXXX");

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Exception in thread "main" java.lang.NoClassDefFoundError: xxxxx (wrong name: xxxxx) error when running progra

By : Ignacio Hernandez
Date : March 29 2020, 07:55 AM
seems to work fine Based on the error message, I can see that Tester is inside a package com.gc.android.market.api
I would expect your console command to be something like

how to replace "?" with in a string with different array values using java/javascript

By : Olaf Tomalka
Date : March 29 2020, 07:55 AM
Hope this helps I want to replace the string "?" with a different array values in the order using javascript , One possible solution is this:
code :
var query = 'SELECT "x0"."NAME", "x0"."DOB", "x0"."SEX", "x0"."CATEGORY"' +
                    'FROM "PERSON" "x0"' +
                    'WHERE ((("x0"."NAME" <> ?) AND ("x0"."NAME" = ?)) AND 

("x0"."NAME" < ?))';

var names = ['borr', 'odin', 'thor'];

for (var i = 0; i < names.length; i++) {
    query = query.replace(/\?/, '"' + names[i] + '"');


How do I change the value in "Xxxxx" format even if user inputs like this "xXXxX"? Java

By : anurag singla
Date : March 29 2020, 07:55 AM
To fix this issue I wrote a small utility for my self. It might help you. I was sure that my string would always have alphabets in it so did not care about any other characters. you might want to modify as per your needs.
code :
int length = "yourstring".length();/// get the length of the string
        String camelCase = removeCharacters.substring(0, 1).toUpperCase();// upper case the first alphabet
        camelCase = camelCase + "yourstring".substring(1, length).toLowerCase();// lowercase all other alphabets

postgres ERROR: insert or update on table "xxxxx" violates foreign key contrain "xxxxx"

By : Jack Tyson
Date : March 29 2020, 07:55 AM
help you fix your problem Did you read through the inheritance docs? Especially the 5.8.1. Caveats section?

The "replace" filter expects an array or "Traversable" as replace values, got "string"

By : Appolia
Date : March 29 2020, 07:55 AM
seems to work fine I had a similar issue - caused I think by an updated version of twig but not 100% sure. I suspect, that somewhere in your twig template (perhaps another not in your examples that is being extended?) you have a replace function like this:
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